Let $X$ be an elliptic curve over $k$, and $P \in X$. We have to prove that for $D=nP$ we have
$$
l(D)=\deg(D)=n
$$
If $D$ satisfies the above equation, and $E$ is a divisor such that $E \geq D$, then we have $l(E)=\deg(E)$ (cf. Fulton, Corollary 8.3.1). Hence it is enough to show that $l(P)=1$.
Clearly $l(P) > 0$, since $k \subset L(P)$. On the other hand, $l(P) > 1$ would imply that there exists $x \in k(X)$ such that $x$ has a simple pole at $P$, and no other poles. But this would imply that the map
$$
x: X \rightarrow \Bbb{P}^1
$$
is an isomorphism, which is a contradiction. Thus $l(P)=1$.
The answer to your first question is yes. As for the second question I would make a minor adjustment but the idea is correct. Please be patient with me as I set things up first.
Notation
Let $\omega_C$ be the sheaf of holomorphic differential forms on $C$. For any vector space $V$ I will denote by $|V|$ the projective space of lines in $V$ and by $\mathbb{P}(V)$ the projective space of codimension 1 planes in $V$, henceforth referred to as hyperplanes.
Throughout I will let $V = H^0(C,\omega_C)$ stand for the global holomorphic differentials on $C$. For a divisor $D$ on $C$ let $V(-D) = H^0(C,\omega_C(-D))$ which is to be viewed as a subspace of $V$ using the natural injection $\omega_C(-D) \hookrightarrow \omega_C$.
Let me write the map $\varphi : C \to \mathbb{P}^{g-1}$ with the proper notation, because as things stand the notation suggests that generators for $V$ have been chosen. The preferred version, as you put it, can be written as $\varphi: C \to \mathbb{P}(V)$ where
$$
p \mapsto V_p := \mathrm{ker}(V \to \omega_C \to \omega_C|_p).
$$
It is clear that $V(-p) \subset V_p$ but Riemann-Roch says that the dimensions match so we get $V(-p) = V_p$.
Basic observations
Given $q \in \mathbb{P}(V)$ we get a hyperplane $H_q \subset |V|$ by duality. Here, $H_q$ parametrizes hyperplanes in $\mathbb{P}(V)$ containing $q$. Given two points $q_1, q_2$ the line between $q_1$ and $q_2$ can be viewed as the intersection of all hyperplanes containing $q_i$'s. This is the dual of $H_{q_1}\cap H_{q_2}$. And similarly for more points. So far this is linear algebra, let us apply it to our specific setting.
Our description of the map $\varphi$ ensures that $H_{\varphi(p)} = |V(-p)|$. Given $p_1, p_2 \in C$ the line between $\varphi(p_1)$ and $\varphi(p_2)$ has dual $|V(-p_1)|\cap|V(-p_2)|$ which is easily seen to be $|V(-p_1-p_2)|$ (Check!). And if $D$ is reduced, this argument is sufficient to conclude that $|V(-D)| \subset |V|$ parametrizes the hyperplanes through $\varphi(D)$.
It is possible, but difficult, to describe the locus of hyperplanes having high order contact at a point $p \in C$ using just geometry. So we go back to doing a little more tautology.
Advanced tautology
If you give me any divisor $D \in |\omega_C| = |V|$ then this gives a line in $H^0(C,\omega_C)$. Pick a section $\sigma$ generating this line. By the standard relations $D = (\sigma)_0$. Now this point $D$ has a dual hyperplane $W_D \subset \mathbb{P}(V)$ consisting of all hyperplanes in $|V|$ containing $D$. I claim that $W_D \cdot \varphi(C) = D$.
Indeed, $W_D$ corresponds to a divisor in the linear system of the tautological line bundle $\mathcal{O}(1)$ which by construction has global sections canonically isomorphic to $V$. Furthermore, the divisor $W_D$ corresponds precisely to the (line generated by) $\sigma \in V$. Now the intersection of $W_D$ with $C$ can be obtained by pulling back $(\mathcal{O}(1),\sigma)$ to $C$. However, the pullback of $\mathcal{O}(1)$ is of course (canonically isomorphic to) $\omega_C$ and the section $\sigma$ maps to $\sigma$ again by canonical identifications. This proves the desired statement.
Wrapping up
So far what we have been doing was largely exploratory. Now let's tackle your question head on. The inclusion $V(-D) \hookrightarrow V$ identifies sections of $\omega_C(-D)$ with sections of $\omega_C$ that vanish on $D$ (this follows from the exact sequence obtained from $\omega_C(-D) \hookrightarrow \omega_C$). Therefore if we take $\sigma \in V(-D)$ then the corresponding hyperplane $H \subset \mathbb{P}(V)$ satisfies $H \cdot C \ge D$. Conversely, any hyperplane that satisfies $H \cdot C \ge D$ corresponds to (the line generated by) a section of $\omega_C$ vanishing on $D$, hence to a section of $\omega_C(-D)$. This answers your first question.
As for the second question, if $d' = \mathrm{deg} D'$ and $d' < g$ then we expect $\varphi(D')$ to span a $d'-1$ dimensional projective subspace. Then your calculation shows that $r(D')$ in fact measures the failure of our expectation. More precisely, $r(D') = d'-1 - \mathrm{dim}(\mathrm{span}(\varphi(D')))$. The space of hyperplanes in $\mathbb{P}^{g-1}$ passing through a $k$-dimensional projective space has dimension $(g-2) - k.$ Then putting these together we get:
$$
r(D) = (g-2) - (d'-1 - r(D')) = (g-1) - (d' - r(D'))
$$
Best Answer
As has already been pointed out, it might be worthwhile learning a bit of algebraic geometry first but I'll try to answer this without going too deep. For most of this I will follow Silverman, as I think elliptic curves make divisors a lot more approachable than abitrary varieties or schemes.
Recall some basics of topology, every compact orientable 2-manifold looks like (is homeomorphic to) an $n$-torus, i.e. either a sphere or a doughnut shape with $n$ holes. This $n$ happens to be equal to the genus of this surface. How do we connect this to elliptic curves though?
Over $\mathbb{C}$, the solution set of an elliptic curve actually looks like a torus topologically (this is called the uniformisation theorem), as we quotient $\mathbb{C}$ by some lattice. Since this is simply a torus with 1 hole, the genus of an elliptic curve is $1$. We can also do this for hyperelliptic curves (although the genus will be different).
On to divisors: Every place of the funtion field of an elliptic curve corresponds to a point on this curve, so we can just think of a divisor as a formal sum of points. The tiny catch is that this sum needs to be finite but that is perfectly fine.
For example, if $E: y^2=x^3+1$ over $\mathbb{C}$, then $Q=(2,3)$ and $R=(0,1)$ are points on $E$. We then have divisors
$D_1 = 2[R], D_2=-2[R]+[Q], D_3=[R]+[Q]$, and these form an abelian group under the obvious addition so $D_1+D_2=D_3$. (Forget about the group law on elliptic curves if you know about it). We also have the concept of the degree of a divisor, which is just the sum of the $n_i$. In the above the degrees of $D_1,D_2$ and $D_3$ are $2,-1$ and $1$ respectively.
Now consider a nonzero function $f$ in the function field of the curve. We say $v_P(f)=n$ if $f$ has a zero of order $n$ at $P$ (note $n$ can be negative if it has a pole instead. Checking for all poles and zeroes of this function, we can construct a divisor out of $f$, which is called $div(f)$.
For example, let $f=x-2$. Then $f=0$ if $x=2$ which occurs for the points $Q=(2,3)$ and $Q'=(2,-3)$ and we can easily see that these are simple zeroes. This means $v_Q(f)=v_{Q'}(f)=1$.
We can also see that $f$ does not have a pole at any point in the affine plane, however, elliptic curves really live in projective space and it turns out that $f$ has a pole of order $2$ at $\infty$. So $v_{\infty}(f)=-2$.
Having found all the zeroes and poles, we now see that the divisor of $f$ is $div(f)=[Q]+[Q']-2[\infty]$. Note that the degree of this is zero and this is always the case for divisors of functions.
For the Riemann Roch space $\mathcal{L}(D)$, we want to find all functions whose divisor is at least the negative of another divisor. For example, if we take $D=[Q]+[Q']$, then we are saying that we are only allowing functions which have at most simple poles at $Q$ and $Q'$ and nowhere else. It is an easy check that this forms a complex vector space.
Adjusting our idea above, $\dfrac{1}{x-2}$ lives in this space (as well as any scale multiple by the vector space structure), but no others do. Hence, we have $1$ basis element so $l(D) \geq 1$. We can also check that $\dfrac{x}{x-2}$ lives in here and that is it so $l(D)=2$. It is useful to note that since all nonconstant functions have poles, then if the degree of $D$ is negative then $l(D)=0$.
Now the Riemann Roch theorem can either be viewed as the definition for the genus, or if you already know this, it can be used to find $l(D)$, up to some correction term.
Now $\mathcal{K}$ is what is known as the canonical divisor and comes from a differential form which I won't attempt to explain here, but has degree $2g-2$.
So in the above, $\mathcal{K}-D$ has degree $-2$ which is negative hence $l(\mathcal{K}-D)=0$ and we can see that the Riemann Roch formula holds.